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Let us say that an Euclidean polyhedral manifold is a manifold that is glued from a finite number of Euclidean polyhedrons by identifying isometrically their co-dimension $1$ faces. Let us assume that no two faces of one polyhedron are glued together and that any two polyhedrons share at most one face.

Let us say that a topological polyhedral manifold is the same thing but this time the faces are identified only by a diffeomorphism, and not necessarily by an isometry. Nevertheless we require that the result of gluing is a manifold.

Question. Suppose now we have a topological polyhedral three manifold. Is it always possible to choose an Euclidean structure on each of its three-dimensional faces (making these faces Euclidean polyhedrons), so that the manifold becomes a Euclidean polyhedral manifold?

Obvious remark. The answer is yes in case of surfaces, one can just say that the surface is glued from regular polygons.

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    $\begingroup$ A direct dimension count seems to show that this problem is underdetermined, so you should be able to construct such a Euclidean polyhedral structure. $\endgroup$
    – Misha
    Sep 26, 2013 at 19:49
  • $\begingroup$ @ Misha I agree, it likely exists, but am afraid of that the dimension count does not suffice. $\endgroup$
    – Sasha Anan'in
    Sep 26, 2013 at 20:15
  • $\begingroup$ @aglearner Please clarify if you assume your Euclidean polyhedra compact or with finitely many faces. Thanks. $\endgroup$
    – Sasha Anan'in
    Sep 27, 2013 at 2:16
  • $\begingroup$ @agleaner Could you please give a definition of Euclidean polyhedra in you question? (Say, are all their faces topological balls? Are they convex in $\Bbb E^3$ (when equipped with an Euclidean structure)?) $\endgroup$
    – Sasha Anan'in
    Sep 27, 2013 at 15:51
  • $\begingroup$ For me a Euclidean polyhedron is just something diffeomorphic to a convex polyhedron in $\mathbb R^3$. $\endgroup$
    – aglearner
    Sep 30, 2013 at 16:36

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